c) Energy Method n k=2 n Fk. Vk + Σ Tk. Wk = k=2 The force due to pressure is: mk .ak Vk + Ik ak. Wk (eq. 1) Fp=(Pressure) x (Surface) = P₁ ×л- k=2 d² 0.038² = 500 × 103 xπ- = 566.7 N and F = -566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp • • Power required: T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: T12= 1935.74 83.776 = 23.1 N.m Problem 3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. The hydraulic cylinder (DB) is designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°; a) Draw clearly the vector loop (DAB). b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular velocity. c) Find the velocity of center of gravity C. d) Using Energy Method, determine the force F required by the hydraulic cylinder (DB) to lift the platform. Neglect the masses of all members. 5 m 2.4 m B 0.5 m A 0.9 m

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Please I want a detailed explanation of the energy method rule placed in the first line and how these numbers appeared to us. I want a detailed explanation please urgent .
c) Energy Method
n
k=2
n
Fk. Vk + Σ Tk. Wk =
k=2
The force due to pressure is:
mk .ak Vk + Ik ak. Wk (eq. 1)
Fp=(Pressure) x (Surface) = P₁ ×л-
k=2
d²
0.038²
= 500 × 103 xπ-
= 566.7 N
and F = -566.7, V = d = 2.9867, ap = 181.02 m/s²
Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp
•
•
Power required:
T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt
Torque required:
T12=
1935.74
83.776
= 23.1 N.m
Transcribed Image Text:c) Energy Method n k=2 n Fk. Vk + Σ Tk. Wk = k=2 The force due to pressure is: mk .ak Vk + Ik ak. Wk (eq. 1) Fp=(Pressure) x (Surface) = P₁ ×л- k=2 d² 0.038² = 500 × 103 xπ- = 566.7 N and F = -566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867+ T12 × W₂ = mp xap xvp • • Power required: T12 W20.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: T12= 1935.74 83.776 = 23.1 N.m
Problem 3
The telescoping arm ABC is used to provide an elevated platform for construction
workers. The workers and the platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. The hydraulic cylinder (DB) is
designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a
constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°;
a) Draw clearly the vector loop (DAB).
b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular
velocity.
c) Find the velocity of center of gravity C.
d) Using Energy Method, determine the force F required by the hydraulic
cylinder (DB) to lift the platform.
Neglect the masses of all members.
5 m
2.4 m
B
0.5 m
A
0.9 m
Transcribed Image Text:Problem 3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. The hydraulic cylinder (DB) is designed to lift the platform so that the arm (ABC) rotates, about the pin A, with a constant angular speed of 0.1 rad/s CW. For the position shown when 0 = 20°; a) Draw clearly the vector loop (DAB). b) Find the velocity of extension of the hydraulic cylinder (DB) and its angular velocity. c) Find the velocity of center of gravity C. d) Using Energy Method, determine the force F required by the hydraulic cylinder (DB) to lift the platform. Neglect the masses of all members. 5 m 2.4 m B 0.5 m A 0.9 m
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